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Commit 308b022

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Aditya1942raklaptudirm
andauthored
merge: Sliding window (#833)
* new algorithm 'Sliding-Window' added with example 'Longest Substring Without Repeating Characters' * new example of sliding window 'Permutation in String' added * chore: add `ignore_words_list` * sliding-window moved into Dynamic-programming directory Co-authored-by: Rak Laptudirm <raklaptudirm@gmail.com>
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‎.github/workflows/Codespell.yml‎

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with:
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# file types to ignore
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skip: "*.json,*.yml,DIRECTORY.md,PermutateString.test.js,SubsequenceRecursive.js"
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ignore_words_list: "ba"
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/**
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* @name The-Sliding-Window Algorithm is primarily used for the problems dealing with linear data structures like Arrays, Lists, Strings etc.
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* These problems can easily be solved using Brute Force techniques which result in quadratic or exponential time complexity.
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* Sliding window technique reduces the required time to linear O(n).
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* @see [The-Sliding-Window](https://www.geeksforgeeks.org/window-sliding-technique/)
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*/
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/**
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* @function LongestSubstringWithoutRepeatingCharacters
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* @description Get the length of the longest substring without repeating characters
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* @param {String} s - The input string
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*/
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export function LongestSubstringWithoutRepeatingCharacters (s) {
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let maxLength = 0
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let start = 0
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let end = 0
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const map = {}
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while (end < s.length) {
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if (map[s[end]] === undefined) {
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map[s[end]] = 1
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maxLength = Math.max(maxLength, end - start + 1)
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end++
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} else {
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while (s[start] !== s[end]) {
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delete map[s[start]]
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start++
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}
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delete map[s[start]]
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start++
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}
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}
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return maxLength
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}
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// Example 1:
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// Input: s = "abcabcbb"
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// Output: 3
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// Explanation: The answer is "abc", with the length of 3.
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// Example 2:
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// Input: s = "bbbbb"
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// Output: 1
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// Explanation: The answer is "b", with the length of 1.
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// Example 3:
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// Input: s = "pwwkew"
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// Output: 3
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// Explanation: The answer is "wke", with the length of 3.
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// Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
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/**
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* @name The-Sliding-Window Algorithm is primarily used for the problems dealing with linear data structures like Arrays, Lists, Strings etc.
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* These problems can easily be solved using Brute Force techniques which result in quadratic or exponential time complexity.
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* Sliding window technique reduces the required time to linear O(n).
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* @see [The-Sliding-Window](https://www.geeksforgeeks.org/window-sliding-technique/)
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*/
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/**
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* @function PermutationinString
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* @description Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.
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* @param {String} s1 - The input string
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* @param {String} s2 - The input string
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* @return {boolean} - Returns true if s2 contains a permutation of s1, or false otherwise.
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*/
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export function PermutationinString (s1, s2) {
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if (s1.length > s2.length) return false
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let start = 0
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let end = s1.length - 1
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const s1Set = SetHash()
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const s2Set = SetHash()
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for (let i = 0; i < s1.length; i++) {
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s1Set[s1[i]]++
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s2Set[s2[i]]++
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}
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if (equals(s1Set, s2Set)) return true
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while (end < s2.length - 1) {
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if (equals(s1Set, s2Set)) return true
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end++
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console.log(s2[start], s2[end], equals(s1Set, s2Set))
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const c1 = s2[start]
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const c2 = s2[end]
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if (s2Set[c1] > 0) s2Set[c1]--
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s2Set[c2]++
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start++
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if (equals(s1Set, s2Set)) return true
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}
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return false
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}
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function equals (a, b) {
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return JSON.stringify(a) === JSON.stringify(b)
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}
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function SetHash () {
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const set = new Set()
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const alphabets = 'abcdefghijklmnopqrstuvwxyz'
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for (let i = 0; i < alphabets.length; i++) {
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set[alphabets[i]] = 0
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}
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return set
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}
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// Example 1:
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// Input: s1 = "ab", s2 = "eidbaooo"
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// Output: true
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// Explanation: s2 contains one permutation of s1 ("ba").
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// Example 2:
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// Input: s1 = "ab", s2 = "eidboaoo"
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// Output: false
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import { LongestSubstringWithoutRepeatingCharacters } from '../LongestSubstringWithoutRepeatingCharacters.js'
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describe('LongestSubstringWithoutRepeatingCharacters', () => {
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it('should return longest substring without repeating characters', () => {
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expect(LongestSubstringWithoutRepeatingCharacters('abcabcbb')).toEqual(3)
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expect(LongestSubstringWithoutRepeatingCharacters('bbbbb')).toEqual(1)
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expect(LongestSubstringWithoutRepeatingCharacters('pwwkew')).toEqual(3)
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expect(LongestSubstringWithoutRepeatingCharacters('a')).toEqual(1)
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expect(LongestSubstringWithoutRepeatingCharacters('')).toEqual(0)
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})
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})
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import { PermutationinString } from '../PermutationinString.js'
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describe('PermutationinString', () => {
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it("should return true if one of s1's permutations is the substring of s2", () => {
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expect(PermutationinString('ab', 'eidbaooo')).toEqual(true)
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expect(PermutationinString('abc', 'bcab')).toEqual(true)
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expect(PermutationinString('ab', 'eidboaoo')).toEqual(false)
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expect(PermutationinString('abc', '')).toEqual(false)
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})
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})

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